Let x = the horizontal distance of travel & y = vertical component of travel t = time
vo = initial velocity of the fish, a = angle of descent as the fish enters the jump. the x & y differential equations for the travel with gravity as the only force acting are:
y"(t) = -g and x(t)" = 0 solving with the initial conditions we get
y(t) = vo(Sin(a)) - (g/2)*t^2 and x(t) = vo(Cos(a))*t
Solving for t from the x equation (2.358/vo) and placing in the y equation , we get
vo = 6.24 m/s